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Q:

# The sum of the squares of three consecutive even numbers is 251. The numbers are?

Q:

The least number of four digits which is divisible by 4, 6, 8 and 10 is ?

 A) 1107 B) 1080 C) 1100 D) 1208

Explanation:

LCM of 4, 6, 8 and 10 = 120
120) 1000 (8
960
------
40

The least number of four digits which is divisible by 4, 6, 8 and 10 => 1000 + 120 - 40 = 1080.

1 2
Q:

Find the least number with which 16200 should be multiplied, to make it a perfect cube ?

 A) 45 B) 48 C) 360 D) 36

Explanation:

16200 = $\inline \fn_jvn \small 2^{3}x3^{4}x5^{2}$

A perfect cube has a property of having the indices of all its prime factors divisible by 3.

Required number = $\inline \fn_jvn \small 3^{2}x5$ = 9x5 = 45.

1 39
Q:

Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is :

 A) 83 B) 89 C) 85 D) 79

Explanation:

Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F of 551 and 1073 = 29;
First number = 551/29 = 19
Third number = 1073/29 = 37.
Required sum = 19 + 29 + 37 = 85.

1 12
Q:

The sum of the two digits of a number is 8. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number ?

 A) 28 B) 19 C) 37 D) 17

Explanation:

Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 8 ----- (1)
(10Q + P) - (10P + Q) = 54
9(Q - P) = 54
(Q - P) = 6 ----- (2)
Solve (1) and (2) P = 1 and Q = 7
The required number is = 17

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Q:

The series of differences between consecutive prime numbers is represented as Cp1, Cp2, Cp3, .... Cpn , Whare Cp1 is the difference between the second and the first prime number. Find the sum of series when n = 23, given that the 23rd prime number is 83 ?

 A) 87 B) 81 C) 83 D) 85

Explanation:

Cp1 = p2 - p1 ,
Cp2 = p3 - p2
..
..
..
Cp23 = p24 - p23
Sum of series = (p2-p1) + (p3-p2) + .....(p23-p22) + (p24-p23)
All terms get cancelled, except p1 = 2 and p24 = 89
So Sum = -p1 + p24
Sum of series = -2 + 89 = 87